Calculate the pH of the above buffer after 10.00 mL of 1.000 M NaOH (0.01000 mol OH1) was added to 1.000 L of thebuffer solution. The new concentrations are [HC2H3O2] = 0.4900 M and [C2H3021] = 0.7100 M. The Ką = 1.60 x 105.%3D%3Deverything is based upon the acid dissolving in water:HC2H3O2+ H20<-> C2H3O21 + H3O*1[Каsolve for [H30*l] then put in the numbersKa [[H30*1] =[H3O*1] =pH = - loga. H30*1b. Он1c. H2C6H606d. HC6H6O61е. HSO31f. SO3?-2g. H2SO3h. HC2H3O2i. C2H3O21j. H2Sk. HS1I. s2m. HXn. X-1О. 1р. 2q. 3r. 0.5000s. 5.5 X 10-5t. 3.5 X 10-4u. 8.00 х 10-5v. 1.75 x 10-6w. 1.11 x 10-8х. 5.5 х 10-2у. О.7000z. 1.60 x 10-5аа. 1.143 х 10-5bb. 0.4900сс. 1.104 х 10-5dd. 0.7100ее. 1.26 х 10-5ff. 4.901gg. 7.93hh. 4.942ii. 4.957

Answered: Image /qna-images/answer/ba0f159c-80a8-4633-9024-80510a072d21.jpg

Calculate the pH of the above buffer after 10.00 mL of 1.000 M NaOH (0.01000 mol OH1) was added to 1.000 L of the buffer solution. The new concentrations are [HC2H3O2] = 0.4900 M and [C2H3O21] = 0.7100 M. The Ka = 1.60 x 105. everything is based upon the acid dissolving in water: HC2H3O2 + H20 <-> C2H3O21 + H3O*1 [ Ka = [ solve for [H30+1] then put in the numbers Ka [ [H3O*l] [H3O*1] = pH = - log a. H30*1 b. Он1 c. H2C6H6O6 d. HC6H6O61 е. HSO31 f. S032 g. H2SO3 h. HC2H3O2 i. C2H3O21 j. H2S k. HS-1 I. s2 m. HX n. X-1 О. 1 р. 2 9. 3 r. 0.5000 s. 5.5 X 10-5 t. 3.5 X 10-4 u. 8.00 x 10-5 v. 1.75 x 10-6 w. 1.11 x 108 x. 5.5 x 10-2 у. О.7000 z. 1.60 x 105 аа. 1.143 х 10-5 bb. 0.4900 сс. 1.104 х 10-5 dd. 0.7100 ее. 1.26 х 10-5 ff. 4.901 gg. 7.93 hh. 4.942 ii. 4.957

Calculate the pH of the above buffer after 10.00 mL of 1.000 M NaOH (0.01000 mol OH1) was added to 1.000 L of the buffer solution. The new concentrations are [HC2H3O2] = 0.4900 M and [C2H3O21] = 0.7100 M. The Ka = 1.60 x 105. everything is based upon the acid dissolving in water: HC2H3O2 + H20 <-> C2H3O21 + H3O1 [ Ka = [ solve for [H30+1] then put in the numbers Ka [ [H3Ol] [H3O1] = pH = - log a. H301 b. Он1 c. H2C6H6O6 d. HC6H6O61 е. HSO31 f. S032 g. H2SO3 h. HC2H3O2 i. C2H3O21 j. H2S k. HS-1 I. s2 m. HX n. X-1 О. 1 р. 2 9. 3 r. 0.5000 s. 5.5 X 10-5 t. 3.5 X 10-4 u. 8.00 x 10-5 v. 1.75 x 10-6 w. 1.11 x 108 x. 5.5 x 10-2 у. О.7000 z. 1.60 x 105 аа. 1.143 х 10-5 bb. 0.4900 сс. 1.104 х 10-5 dd. 0.7100 ее. 1.26 х 10-5 ff. 4.901 gg. 7.93 hh. 4.942 ii. 4.957

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

propionic acid, ?a=1.34×10−5,Ka=1.34×10−5, 3.00 M sodium propionate, CH3CH2COONa 1. The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on this information, what mass of solid conjugate base should the student weigh out to make the buffer solution with a pH of 5.00? ( in g) 2. Based on this information, what volume of acid should the student measure to make the buffer solution? (in ml)
HCN is a weak acid with a Ka = 4.90 x 10 10. A 50.00 mL sample of this HCN solution (0.250M) is titrated with 0.500 M NaOH. What is the pH after 15 mL of NaOH is added?
Use the References to access important values if needed for this question. A buffer solution is made that is 0.379 M in HCN and 0.379 M in NaCN. If Ka for HCN is 4.00 x 10-¹0, what is the pH of the buffer solution? pH = Write the net ionic equation for the reaction that occurs when 0.090 mol HNO3 is added to 1.00 L of the buffer solution. (Use the lowest possible coefficients. Omit states of matter. Use H3O+ instead of H+) + Submit Answer ->> Retry Entire Group + 9 more group attempts remaining
A 1.50 L buffer solution consists of 0.280 M propanoic acid and 0.139 M sodium propanoate. Calculate the pH of the solution following the addition of 0.065 mol HCl. Assume that any contribution of the HCl to the volume of the solution is negligible. The Ka of propanoic acid is 1.34×10^−5.
An analytical chemist is titrating 75.3 mL of a 0.2100M solution of cyanic acid (HCNO) with a 0.8800M solution of NaOH. The p K of cyanic acid is 3.46. Calculate the pH of the acid solution after the chemist has added 20.9 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places. pH = 0
a An analytical chemist is titrating 86.1 mL of a 1.100M solution of hydrazoic acid (HN3) with a 0.7600M solution of KOH. The p K of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 74.4 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places.
1) A buffer solution is 0.337 M in H2CO3 and 0.239 M in KHCO3 . If Ka1 for H2CO3 is 4.2x10^-7 , what is the pH of this buffer solution? pH = 2) An aqueous solution contains 0.302 M hypochlorous acid. How many mL of 0.231 M potassium hydroxide would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 7.540? 3) 36.5 mL sample of a 0.310 M aqueous hypochlorous acid solution is titrated with a 0.441 M aqueous barium hydroxide solution. What is the pH after 8.80 mL of base have been added? Ka for HClO is 3.5x10^-8 . pH =
You start with 100.0 mL of 0.20 M CH3COOH in a flask. The Ka of CH3COOH is 1.8 x 10-5. You add 10.0 mL of 0.20 M NaOH. What is the pH of the solution after the NaOH is added? TRUE or FALSE: The solution produced by mixing the reagents in the question above is a buffer.
Calculate the expected pH of the buffer.
An analytical chemist is titrating 145.0 mL of a 0.7900M solution of dimethylamine ((CH) NH) with a 0.8700M solution of HNO3. The p K, of dimethylamine is 3.27. Calculate the pH of the base solution after the chemist has added 93.9 mL of the HNO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places. pH =
Determine the pH at the point in the titration of 40.0 mL of 0.200 M HC4H,O2 with 0.100 M Sr(OH)2 after 100.0 mL of the strong base has been added. The value of Ka for HC4H;O2 is 1.5 × 10-5 1 2 NEXT > Use the table below to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of liquid water in the reaction. HC4H;O2(aq) + OH (aq) H20(1) + C4H,O2 (aq) Before (mol) |Change Imol) After | (mol) RESET 0.100 0.200 +x 0.100 + x 0.100 - x -x 0.200 + x 0.200 - x 2.00 x 103 -2.00 x 103 8.00 x 103 -8.00 х 10"3 1.00 x 102 -1.00 x 102 1.20 х 10"2 -1.20 x 102 2.00 x 10-2 -2.00 x 102